Answer:
16.933g approximately 17.0 grams
Explanation:
From the simple promotions and given the same compound ascorbic acid (vitamin C)
In the laboratory synthesised ascorbic acid
Mass of carbon = 30.0g
Mas of Oxygen = 40.0g
That is the mass of Oxygen per unit mass of Carbon
Per gram of Carbon we have
(30.0g Carbon)÷30 combines with (40.0g of Oxygen)÷30
That is 4/3g of Oxygen per gram of Carbon
Hence the mass of Oxygen compound that combines with 12.7g of Carbonin natural occurring ascorbic acid (vitamin C) is = 4/3×12.7 = 16.933g approximately 17.0g
3.0 × 10¹¹ RBC's (or) 3E11 RBC's
Solution:
Step 1: Convert mm³ into L;
As,
1 mm³ = 1.0 × 10⁻⁶ Liters
So,
0.1 mm³ = X Liters
Solving for X,
X = (0.1 mm³ × 1.0 × 10⁻⁶ Liters) ÷ 1 mm³
X = 1.0 × 10⁻⁷ Liters
Step 2: Calculate No. of RBC's in 5 Liter Blood:
As given
1.0 × 10⁻⁷ Liters Blood contains = 6000 RBC's
So,
5.0 Liters of Blood will contain = X RBC's
Solving for X,
X = (5.0 Liters × 6000 RBC's) ÷ 1.0 × 10⁻⁷ Liters
X = 3.0 × 10¹¹ RBC's
Or,
X = 3E11 RBC's
More unstable an electron configuration , the more reactive an atom will become.
How electron configuration influences the chemical behavior of an atom?
This is happen generally, If we look at the Group 1 elements in the periodic table, they are all highly reactive as they have 1 electron in their outermost shells - an unstable configuration in terms of energy.
Also, the noble gases in Group 8 in the periodic table are 'inert' that means they don't react (or more correctly, have an incredibly low reactivity). This is because they have 8 electrons in their outermost shell and thus have no need to acquire or lose electrons to possess a stable electron configuration.
Hence, electron configuration influences the chemical behavior of an atom.
learn more about electronic configuration here :
brainly.com/question/26084288
#SPJ4
Answer : The correct option is, (b) +115 J/mol.K
Explanation :
Formula used :
where,
= change in entropy
= change in enthalpy of vaporization = 40.5 kJ/mol
= boiling point temperature = 352 K
Now put all the given values in the above formula, we get:
Therefore, the standard entropy of vaporization of ethanol at its boiling point is +115 J/mol.K
